What are the solutions to the equation (x-2)^2+(4/3 x+4/3-4)^2=16 ?

The solutions to the equation (x-2)^2+(4/3 x+4/3-4)^2=16 are x= 22/5 ,x=-2/5

Find the zeros of (x-2)^2+(4/3 x+4/3-4)^2=16

The zeros of (x-2)^2+(4/3 x+4/3-4)^2=16 are x= 22/5 ,x=-2/5

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(x - 2)^{2} + (\frac{4}{3} x + \frac{4}{3} - 4)^{2} = 16

x = \frac{22}{5}, x = - \frac{2}{5}

Decimal Notation

x = 4.4, x = - 0.4

Solution steps

(x - 2)^{2} + (\frac{4}{3} x + \frac{4}{3} - 4)^{2} = 16

Expand

(x - 2)^{2} + (\frac{4}{3} x + \frac{4}{3} - 4)^{2} : \frac{100}{9} + \frac{25 x ^{2}}{9} - \frac{100 x}{9}

\frac{100}{9} + \frac{25 x ^{2}}{9} - \frac{100 x}{9} = 16

Move

16

to the left side

\frac{100 + 25 x ^{2} - 100 x}{9} - 16 = 0

Write in the standard form

a x^{2} + b x + c = 0

\frac{25 x ^{2}}{9} - \frac{100 x}{9} + \frac{100}{9} - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - \frac{100}{9} ) \pm ( - \frac{100}{9} ) ^{2} - 4 \cdot \frac{25}{9} ( \frac{100}{9} - 16 )}{2 \cdot \frac{25}{9}}

(- \frac{100}{9})^{2} - 4 \cdot \frac{25}{9} (\frac{100}{9} - 16) = \frac{40}{3}

x_{1, 2} = \frac{- ( - \frac{100}{9} ) \pm \frac{40}{3}}{2 \cdot \frac{25}{9}}

Separate the solutions

x_{1} = \frac{- ( - \frac{100}{9} ) + \frac{40}{3}}{2 \cdot \frac{25}{9}}, x_{2} = \frac{- ( - \frac{100}{9} ) - \frac{40}{3}}{2 \cdot \frac{25}{9}}

x = \frac{- ( - \frac{100}{9} ) + \frac{40}{3}}{2 \frac{25}{9}} : \frac{22}{5}

x = \frac{- ( - \frac{100}{9} ) - \frac{40}{3}}{2 \frac{25}{9}} : - \frac{2}{5}

The solutions to the quadratic equation are:

x = \frac{22}{5}, x = - \frac{2}{5}

(x + 3) (x + 5) = 48

4 p^{2} - 4 = 0

x^{2} + 4 x + x = 0

9 v^{2} + 30 v = - 25

7 f^{2} = f + 1

What are the solutions to the equation (x-2)^2+(4/3 x+4/3-4)^2=16 ?
The solutions to the equation (x-2)^2+(4/3 x+4/3-4)^2=16 are x= 22/5 ,x=-2/5
Find the zeros of (x-2)^2+(4/3 x+4/3-4)^2=16
The zeros of (x-2)^2+(4/3 x+4/3-4)^2=16 are x= 22/5 ,x=-2/5